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Calculus learning

Improper Integrals: Infinite Bounds, Singularities and Convergence

An improper integral contains an infinite interval or an unbounded integrand. The integral sign alone does not settle convergence. Work on a finite truncated interval first, then take the appropriate one-sided limit; when there are two problematic ends, evaluate them separately.

Direct answer

Replace each infinite endpoint or singular endpoint by its own limit. Every required limit must converge.

Split at every interior singularity. A Cauchy principal value is not the ordinary improper integral.

Improper Integrals worked method

For ∫₁^∞ x⁻² dx, first integrate from 1 to R: 1−1/R. Its limit as R→∞ is 1, so the improper integral converges.

For ∫₁^∞ 1/x dx, the truncated result is ln R. It increases without bound; the integral diverges even though the integrand tends to zero.

At a finite singular endpoint, ∫₀¹ x⁻¹ᐟ² dx is the limit of 2−2√ε as ε→0⁺. The finite limit is 2.

For ∫₋₁¹ 1/x dx, split at zero. The left integral tends to −∞ and the right to +∞. Cancelling them in a symmetric truncation produces a principal value, not a convergent ordinary integral.

For the p-test, ∫₁^∞ x⁻ᵖ dx converges exactly when p>1; ∫₀¹ x⁻ᵖ dx converges exactly when p<1. The locations of the singularity change the condition.

1
∫1Rx−2 dx\int_{1}^{R} x^{-2}\,\mathrm{d}x
2
∫1Rx−1 dx\int_{1}^{R} x^{-1}\,\mathrm{d}x
3
lim ε→0⁺ ∫ε¹ x⁻¹ᐟ² dx = 2

Worked examples

∫1∞1x2 dx\int_{1}^{\infty} \frac{1}{x^{2}}\,\mathrm{d}x
11

Decay is fast enough.

∫1∞1x dx\int_{1}^{\infty} \frac{1}{x}\,\mathrm{d}x
Diverges

Going to zero is necessary here but not sufficient.

∫₀¹ 1/√x dx
22

A finite endpoint may still be improper.

Common mistakes

  • • Treating +∞ as an ordinary substitution value.
  • • Taking a single symmetric limit across a pole.
  • • Claiming convergence merely because f(x)→0.

Try the supported calculator preset

The preset opens a supported expression with its variable and conditions. Teaching examples elsewhere in this guide are reference explanations, not claims that the engine supports every method.

This example computes the finite truncation [1,10], not the full infinite-bound integral. The convergence argument remains visible in the lesson.

積分:1x2\frac{1}{x^{2}}
Open the calculator without a preset: 積分

References

Frequently asked questions

Does a numerical approximation establish convergence?+

No. A finite truncation cannot prove that an infinite tail has a finite total.

Why can two infinite terms not cancel?+

The ordinary integral requires each side separately to exist finitely.

Does the calculator support all improper integrals?+

No. Its infinite-bound rules are limited. The button here opens a finite truncation only.

Conclusion

Improper Integrals starts with the stated domain and conditions. Recheck those before carrying a worked example into a different problem.

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